Unzipping file in android/kotlin

UnzipUtils.kt

import java.io.*
import java.util.zip.ZipFile


/**
 * UnzipUtils class extracts files and sub-directories of a standard zip file to
 * a destination directory.
 *
 */
object UnzipUtils {
    /**
     * @param zipFilePath
     * @param destDirectory
     * @throws IOException
     */
    @Throws(IOException::class)
    fun unzip(zipFilePath: File, destDirectory: String) {
        
          File(destDirectory).run {
                if (!exists()) {
                    mkdirs()
                }
        }
            
        ZipFile(zipFilePath).use { zip ->

            zip.entries().asSequence().forEach { entry ->

                zip.getInputStream(entry).use { input ->


                        val filePath = destDirectory + File.separator + entry.name

                        if (!entry.isDirectory) {
                            // if the entry is a file, extracts it
                            extractFile(input, filePath)
                        } else {
                            // if the entry is a directory, make the directory
                            val dir = File(filePath)
                            dir.mkdir()
                        }

                }

            }
        }
    }

    /**
     * Extracts a zip entry (file entry)
     * @param inputStream
     * @param destFilePath
     * @throws IOException
     */
    @Throws(IOException::class)
    private fun extractFile(inputStream: InputStream, destFilePath: String) {
        val bos = BufferedOutputStream(FileOutputStream(destFilePath))
        val bytesIn = ByteArray(BUFFER_SIZE)
        var read: Int
        while (inputStream.read(bytesIn).also { read = it } != -1) {
            bos.write(bytesIn, 0, read)
        }
        bos.close()
    }

    /**
     * Size of the buffer to read/write data
     */
    private const val BUFFER_SIZE = 4096

}


/**

Copyright 2020 Nitin Prakash

Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at

   http://www.apache.org/licenses/LICENSE-2.0

Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
*/

val folder = applicationContext.getExternalFilesDir(test.zip)
val destinationPath = "/storage/emulated/0/advertisments/unzip"
SystemUtils().unzip(folder!!, destinationPath)

why always get error Caused by: java.util.zip.ZipException: error in opening zip file?

Ps: you may want to change it as

    @Throws(IOException::class)
    infix fun File.unzip(destDirectory: String) {
        File(destDirectory).run {
            if (!exists())
                mkdirs()
        }
        ZipFile(this).use { zip ->
            zip.entries().asSequence().forEach { entry ->
                zip.getInputStream(entry).use { input ->
                    val filePath = destDirectory / entry.name
                    if (entry.isDirectory) // if the entry is a directory, make the directory
                        File(filePath).mkdir()
                    else // if the entry is a file, extracts it
                        input.copyTo(FileOutputStream(filePath))
                }
            }
        }
    }

                val filePath = destDirectory / entry.name

To avoid ambiguity of String and URI
val filePath: String = "${destDirectory}/${entry.name}"